Quartic equation with integer solutions I am trying to find the solutions of the following quartic equation given that all the solutions are integers.
$$x^4+22x^3+172x^2+552x+576=0$$
Below is the original phrasing of the problem with hints building up to this equation. I can prove all of the results it asks for before the final part of the question, but I am struggling to actually find the solutions to the equation and require help with explaining the process as well.



I then know that:
(1) $k_1k_2k_3k_4 = 576 = (1)(2^6)(3^2)$
(2) $(k_1+1)(k_2+1)(k_3+1)(k_4+1) = 1323=(1)(3^3)(7^2)$
(3) $(k_1-1)(k_2-1)(k_3-1)(k_4-1) = 175 = (1)(5^2)(7)$
However, I do not know how to proceed from here and the solution to this problem didn't explain in enough detail for me to either understand the solution, or the approach.
 A: This is an attempt to show how the method suggested in the question can be progressed.
Look at $1323$ with factors $3^3\times 7^2$. The size of $7$ will restrict the possibilities to investigate more quickly than the smaller primes. We have $576=24^2$
The available multiples of $7$ are $7, 21, 49, 63, 147$ (others are too large) giving possible factors of $576$ which differ by $1$ (can't tell the sign at this stage).
So the possible factors of $576$ are $6 , 8; 20, 22; 48, 50; 62, 64; 146, 148$ and the only ones which are actually factors are $6, 8, 48$
And the possible factors of $175$ differ by $2$ (in the same direction as the factors of $576$) so can be $5, 9; 47$ and this time only $5$ is possible, with two factors to match the two $7$s. 
This means the factors we have for $576$ have numerical value $6$, and in fact we can tell that the roots are $-6$ by attention to signs [NB]. Then there are lots of ways to finish - having two solutions, we can, for example, identify and solve the quadratic for the remaining roots. Or alternatively the remaining factor $7$ from $175$ gives $-8$ and the final factor has to be $1$ which gives $-2$ (signs to be allocated with care, but since coefficients are all positive, roots must all be negative see NB below).
NB The $k_i$ are the negatives of the roots so the $k_i$ here would be $6$ with the factors $(x+6)^2$ and the double root $x=-6$. And then $8$ and $2$ similarly.
A: Observe that the given equations imply that $k_1,k_2,k_3$ and $k_4$ should be non-zero even integers. Now consider $$(k_1-1)(k_2-1)(k_3-1)(k_4-1)=1\times 5^2 \times 7.$$
Since, $k_i$'s are integers, one of the $k_i-1$ must be $1$ ($k_i-1 \neq -1$ as $k_i \neq 0$). Let $k_1-1=1 \implies \color{red}{k_1 =2}$. 
We are now left with $$(k_2-1)(k_3-1)(k_4-1)= 5^2 \times 7.$$
Since we have $5^2$ on the RHS, at least one of the $k_i-1$ (say $k_2-1$) cannot be a multiple of $5$. Consequently, $k_2-1 = 1$ or $k_2-1=\pm 7$ (i.e., $k_2 =2,8$ or $-6$). But $k_2 + 1$ must be a multiple of $3$ or $7$ (due to the second constraint). So $k_2 \neq -6$. Therefore, $k_2=2$ or $k_2=8$.
Now suppose $k_2=2$. Then the constraints simplify to
$$k_3 k_4 = 2^4 \times 3^2$$
$$(k_3+1)(k_4+1) = 3 \times 7^2$$
$$(k_3-1)(k_4-1) = 5^2 \times 7$$
It can be verified that these equations do not have any solution. 
Therefore, we are left with $\color{red}{k_2 = 8}$ only. In that case, the equations simplify to
$$k_3 k_4 = 2^2 \times 3^2$$
$$(k_3+1)(k_4+1) = 7^2$$
$$(k_3-1)(k_4-1) = 5^2$$
These equations can easily be solved to result in $\color{red}{k_3 =k_4 = 6}$.
A: Lemma: 
Let 
$ 
P(x)= 
a_nx^n+ 
a_{n-1}x^{n-1}+ 
... 
+ 
a_1x+ 
a_0 
$ 
be a polynomail with integral coefficeint;
i.e. $a_i \in \mathbb{Z}$ and $a_n\neq 0$.
Let $\alpha=\dfrac{r}{s}$ be a rational root of this equation, 
with $\gcd(r,s)=1$;
then we must have: $s \mid a_n$ and $r \mid a_0$ .
Proof: Let $\alpha=\dfrac{r}{s}$ be a rational root of $P(x)$, 
with $\gcd(r,s)=1$; then we have: 
$$ 
0= 
P(\alpha)= 
a_n\alpha^n+ 
a_{n-1}\alpha^{n-1}+ 
... 
+ 
a_1\alpha+ 
a_0 
\Longrightarrow 
\\ 
\ \ \ \ \ 
\ \ \ \ \ 
0= 
a_n(\dfrac{r}{s})^n+ 
a_{n-1}(\dfrac{r}{s})^{n-1}+ 
... 
+ 
a_1(\dfrac{r}{s})+ 
a_0 
\Longrightarrow 
\\
\ \ \ \ \ 
\ \ \ \ \ 
\ \ \ \ \ 
\ \ \ \ \ 
\ \ \ \ \ 
\ 
0= 
a_nr^n+ 
a_{n-1}r^{n-1}s+ 
... 
+ 
a_1rs^{n-1}+ 
a_0s^n 
\ \ \ \ \ \ \ 
\star
$$ 
note that $r$ divides the LHS of $\star$; so it must divides the RHS;
also notice that $r$ divides all terms except $a_0s^n$; 
so it must divides $a_0s^n$;
by Euclid's lemma we can conclude that $r|a_0$. 
note that $s$ divides the LHS of $\star$; so it must divides the RHS;
also notice that $s$ divides all terms except $a_nr^n$; 
so it must divides $a_nr^n$;
by Euclid's lemma we can conclude that $s|a_n$. 

In the special case of this question we know $576=2^63^2$, so we must have: 
$$s|1 \ \ \ \ \text{and} \ \ \ \ r|2^63^2 $$
so it only suffices to check all the integers $\pm d$; 
where $d$ is a positive divisor of $576$;
one can check by hand that the only posibilities are $-2$ and $-6$. 
A: For $$f(x) = x^4+22x^3+172x^2+552x+576,$$
notice that if $x\geq0$ then $f(x) \geq 576 > 0,$
so all roots of the equation $f(x) = 0$ are negative.
Since the roots are $-k_1, -k_2, -k_3, -k_4,$
this tells us that $k_1, k_2, k_3, k_4$ are all positive.
That leaves only a few possible values of $k_i$ that could occur
in the equation
$$(k_1-1)(k_2-1)(k_3-1)(k_4-1) = 175 = 5^2 \cdot 7.$$
The factors of $175$ are $1, 5, 5^2 = 35, 7, 5\cdot7 = 35,$ 
and $5^2 \cdot 7 = 175.$
(I chose this equation to start with because $175$ has fewer factors to consider than either $576$ or $1323$.)
Since each factor $k_i - 1$ must be a factor of $175,$
the possible values of any $k_i$ can only be among the numbers
$2, 6, 26, 8, 36,$ and $176.$
From 
$$k_1k_2k_3k_4 = 576 = 2^6 \cdot 3^2, \tag1$$
we know any of the $k_i$ can have only $2$ or $3$ as prime factors.
So $k_i \neq 26 = 2 \cdot 13$ and $k_i \neq 176 = 16 \cdot 11.$
From
$$(k_1+1)(k_2+1)(k_3+1)(k_4+1) = 1323 = 3^3 \cdot 7^2,$$
we know $k_i + 1$ must be divisible by $3$ or by $7.$
So $k_i + 1 \neq 37,$ and therefore $k_i \neq 36.$
Therefore the only possible values of any of the $k_i$ are $2,$ $6,$ or $8.$
But Equation $(1)$ implies either that one of the $k_i$ is divisible by $9$
(which we now know cannot be true) or that two of the $k_i$ are each divisible by $3.$ If $k_i$ is divisible by $3,$ its only possible value is $6$ (since neither $2$ and $8$ is divisible by $3$).
Without loss of generality, we can set $k_1 = k_2 = 6.$
Dividing both sides of Equation $(1)$ by $k_1k_2 = 36,$
we now have $k_3k_4 = 2^4.$ Each of $k_3$ or $k_4$ can then only be $2$ or $8$ (since $6$ is not a power of $2$), but if one is $2$ then the other is $8$ and vice versa. So without loss of generality we can set $k_3 = 2$ and
$k_4 = 8.$
The four roots therefore are
$-k_1 = -6,$ $-k_2 = -6,$ $-k_3 = -2,$ and $-k_4 = -8.$
In ascending order they are $-8, -6, -6, -2.$
A: Your equations (2) and (3) seem an over-complicated way to attack the problem.
The fact that $k_1k_2k_3k_4 = 576 = 2^6 3^2$ is useful. So is the fact that $k_1 + k_2 + k_3 + k_4 = -22$, which you can prove in a similar fashion. (Note, both of these are general results for any polynomial).
For a third useful result, by inspection the equation has no positive roots, since the sign of every term is positive. This is a particular case of "Descartes' rule of signs."
Those three facts are usually enough to guess the values of $k_i$. If not, it may be simpler to try to find two quadratic factors, rather than four linear ones - the fact that the quadratic factors are not unique can often make it easier to guess one pair of them.
For this specific polynomial, this means that $0< k_i < 22$ for each $k_i$. The only possible integer values in that range which are factors of $576$ are $1, 2, 3, 4, 6, 8, 9, 12, 16, 18$.  
By trial and error $1$ does not give a root, but $2$ does. That eliminates $18$, because $18+2+2+2 > 22$. 
So to get the $3^2$ factor in $576$, we need to include some of $3, 6, 9$, or $12$. Of those four, the only root is $6$. 
So we must have  $k_1 = k_2 = 6$ and we already know $k_3 = 2$ gives a root. So $k_4 = 8$.
A: We have
$\displaystyle 576=2^6 \cdot 3^2$
The number $\displaystyle 3$ will be factor of at most two roots, so at least two roots will be powers of $\displaystyle 2$. So let's try the candidate roots
$\displaystyle \pm 2, \pm 4, \pm 8, \pm 16, \pm 32, \pm 64$
So we find two roots:
$\displaystyle x_1=-2, x_2=-8$
We will have now
$\displaystyle 22=-(-2)-(-8)-x_3-x_4 \Rightarrow$
$\displaystyle 12=-x_3-x_4$
and
$\displaystyle \frac{576}{(-2)(-8)}=36=x_3 x_4$
From the last two systems of equations we are easily led to the equation
$\displaystyle x_{3,4}^2+12x_{3,4}-36=0$
which is easily solved and gives us
$\displaystyle x_3=x_4=-6$
